Problem: The equation of a circle $C$ is $x^2+y^2+4x+12y+24 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+4x) + (y^2+12y) = -24$ $(x^2+4x+4) + (y^2+12y+36) = -24 + 4 + 36$ $(x+2)^{2} + (y+6)^{2} = 16 = 4^2$ Thus, $(h, k) = (-2, -6)$ and $r = 4$.